$\displaystyle \int tan^{-1}u du $
is equal to
$\displaystyle utan^{-1}u-\frac{1}{2} ln(1+u^{2}) $
can somebody explain to me how this is true?
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$\displaystyle \int tan^{-1}u du $
is equal to
$\displaystyle utan^{-1}u-\frac{1}{2} ln(1+u^{2}) $
can somebody explain to me how this is true?
Did you differentiate the answer?Quote:
Originally Posted by akhayoon
Do it!
That shows you how it works.
well thats the thing I haven't been taught these
it's just given to me as a formula...but I would like to understand how it happened so that way I dont have to memorize it...
thanks
Sorry to say, but I find that notion ridiculous.Quote:
Originally Posted by akhayoon
well..I really don't care you how you find that notion...all I really want is a hint on how to begin solving that problem really, which is what I always get when I have a question on this forum.
Quote:
Originally Posted by akhayoon
Here's your hint, as Plato suggested:Quote:
Originally Posted by Plato
$\displaystyle \frac{d}{du} \left ( u~tan^{-1}(u)-\frac{1}{2} ln(1+u^{2}) \right )$
$\displaystyle = tan^{-1}(u) + \frac{u}{u^2 + 1} - \frac{1}{2} \left ( \frac{1}{1 + u^2} \cdot 2u \right )$
What does this tell you?
-Dan
the responses given seem to me to be geared towards verifying that the claim is true, rather than saying why it's true, or how it works. i believe what the poster was after is actually integrating arctan to get the result shown. in that case, we use integration by parts with $\displaystyle u = \arctan x$ and $\displaystyle dv = 1$ (you do know what i am referring to when i say $\displaystyle u \mbox{ and } dv$, right akhayoon?)
A similar integration by parts reasoning is useful here: take the following function $\displaystyle f(u)=u\arctan u.$ Contemplate its derivative and integrate, the rest follows.
